Permutations¶

class
sympy.combinatorics.permutations.
Permutation
[source]¶ A permutation, alternatively known as an ‘arrangement number’ or ‘ordering’ is an arrangement of the elements of an ordered list into a onetoone mapping with itself. The permutation of a given arrangement is given by indicating the positions of the elements after rearrangement [R30]. For example, if one started with elements [x, y, a, b] (in that order) and they were reordered as [x, y, b, a] then the permutation would be [0, 1, 3, 2]. Notice that (in SymPy) the first element is always referred to as 0 and the permutation uses the indices of the elements in the original ordering, not the elements (a, b, etc…) themselves.
>>> from sympy.combinatorics import Permutation >>> Permutation.print_cyclic = False
See also
References
[R29] Skiena, S. ‘Permutations.’ 1.1 in Implementing Discrete Mathematics Combinatorics and Graph Theory with Mathematica. Reading, MA: AddisonWesley, pp. 316, 1990. [R30] (1, 2) Knuth, D. E. The Art of Computer Programming, Vol. 4: Combinatorial Algorithms, 1st ed. Reading, MA: AddisonWesley, 2011. [R31] Wendy Myrvold and Frank Ruskey. 2001. Ranking and unranking permutations in linear time. Inf. Process. Lett. 79, 6 (September 2001), 281284. DOI=10.1016/S00200190(01)001417 [R32] D. L. Kreher, D. R. Stinson ‘Combinatorial Algorithms’ CRC Press, 1999 [R33] Graham, R. L.; Knuth, D. E.; and Patashnik, O. Concrete Mathematics: A Foundation for Computer Science, 2nd ed. Reading, MA: AddisonWesley, 1994. [R34] (1, 2) http://en.wikipedia.org/wiki/Permutation#Product_and_inverse [R35] http://en.wikipedia.org/wiki/Lehmer_code Permutations Notation
Permutations are commonly represented in disjoint cycle or array forms.
Array Notation And 2line Form
In the 2line form, the elements and their final positions are shown as a matrix with 2 rows:
[0 1 2 … n1] [p(0) p(1) p(2) … p(n1)]
Since the first line is always range(n), where n is the size of p, it is sufficient to represent the permutation by the second line, referred to as the “array form” of the permutation. This is entered in brackets as the argument to the Permutation class:
>>> p = Permutation([0, 2, 1]); p Permutation([0, 2, 1])
Given i in range(p.size), the permutation maps i to i^p
>>> [i^p for i in range(p.size)] [0, 2, 1]
The composite of two permutations p*q means first apply p, then q, so i^(p*q) = (i^p)^q which is i^p^q according to Python precedence rules:
>>> q = Permutation([2, 1, 0]) >>> [i^p^q for i in range(3)] [2, 0, 1] >>> [i^(p*q) for i in range(3)] [2, 0, 1]
One can use also the notation p(i) = i^p, but then the composition rule is (p*q)(i) = q(p(i)), not p(q(i)):
>>> [(p*q)(i) for i in range(p.size)] [2, 0, 1] >>> [q(p(i)) for i in range(p.size)] [2, 0, 1] >>> [p(q(i)) for i in range(p.size)] [1, 2, 0]
Disjoint Cycle Notation
In disjoint cycle notation, only the elements that have shifted are indicated. In the above case, the 2 and 1 switched places. This can be entered in two ways:
>>> Permutation(1, 2) == Permutation([[1, 2]]) == p True
Only the relative ordering of elements in a cycle matter:
>>> Permutation(1,2,3) == Permutation(2,3,1) == Permutation(3,1,2) True
The disjoint cycle notation is convenient when representing permutations that have several cycles in them:
>>> Permutation(1, 2)(3, 5) == Permutation([[1, 2], [3, 5]]) True
It also provides some economy in entry when computing products of permutations that are written in disjoint cycle notation:
>>> Permutation(1, 2)(1, 3)(2, 3) Permutation([0, 3, 2, 1]) >>> _ == Permutation([[1, 2]])*Permutation([[1, 3]])*Permutation([[2, 3]]) True
Caution: when the cycles have common elements between them then the order in which the permutations are applied matters. The convention is that the permutations are applied from right to left. In the following, the transposition of elements 2 and 3 is followed by the transposition of elements 1 and 2:
>>> Permutation(1, 2)(2, 3) == Permutation([(1, 2), (2, 3)]) True >>> Permutation(1, 2)(2, 3).list() [0, 3, 1, 2]
If the first and second elements had been swapped first, followed by the swapping of the second and third, the result would have been [0, 2, 3, 1]. If, for some reason, you want to apply the cycles in the order they are entered, you can simply reverse the order of cycles:
>>> Permutation([(1, 2), (2, 3)][::1]).list() [0, 2, 3, 1]
Entering a singleton in a permutation is a way to indicate the size of the permutation. The
size
keyword can also be used.Arrayform entry:
>>> Permutation([[1, 2], [9]]) Permutation([0, 2, 1], size=10) >>> Permutation([[1, 2]], size=10) Permutation([0, 2, 1], size=10)
Cyclicform entry:
>>> Permutation(1, 2, size=10) Permutation([0, 2, 1], size=10) >>> Permutation(9)(1, 2) Permutation([0, 2, 1], size=10)
Caution: no singleton containing an element larger than the largest in any previous cycle can be entered. This is an important difference in how Permutation and Cycle handle the __call__ syntax. A singleton argument at the start of a Permutation performs instantiation of the Permutation and is permitted:
>>> Permutation(5) Permutation([], size=6)
A singleton entered after instantiation is a call to the permutation – a function call – and if the argument is out of range it will trigger an error. For this reason, it is better to start the cycle with the singleton:
The following fails because there is is no element 3:
>>> Permutation(1, 2)(3) Traceback (most recent call last): ... IndexError: list index out of range
This is ok: only the call to an out of range singleton is prohibited; otherwise the permutation autosizes:
>>> Permutation(3)(1, 2) Permutation([0, 2, 1, 3]) >>> Permutation(1, 2)(3, 4) == Permutation(3, 4)(1, 2) True
Equality Testing
The array forms must be the same in order for permutations to be equal:
>>> Permutation([1, 0, 2, 3]) == Permutation([1, 0]) False
Identity Permutation
The identity permutation is a permutation in which no element is out of place. It can be entered in a variety of ways. All the following create an identity permutation of size 4:
>>> I = Permutation([0, 1, 2, 3]) >>> all(p == I for p in [ ... Permutation(3), ... Permutation(range(4)), ... Permutation([], size=4), ... Permutation(size=4)]) True
Watch out for entering the range inside a set of brackets (which is cycle notation):
>>> I == Permutation([range(4)]) False
Permutation Printing
There are a few things to note about how Permutations are printed.
1) If you prefer one form (array or cycle) over another, you can set that with the print_cyclic flag.
>>> Permutation(1, 2)(4, 5)(3, 4) Permutation([0, 2, 1, 4, 5, 3]) >>> p = _
>>> Permutation.print_cyclic = True >>> p (1 2)(3 4 5) >>> Permutation.print_cyclic = False
2) Regardless of the setting, a list of elements in the array for cyclic form can be obtained and either of those can be copied and supplied as the argument to Permutation:
>>> p.array_form [0, 2, 1, 4, 5, 3] >>> p.cyclic_form [[1, 2], [3, 4, 5]] >>> Permutation(_) == p True
3) Printing is economical in that as little as possible is printed while retaining all information about the size of the permutation:
>>> Permutation([1, 0, 2, 3]) Permutation([1, 0, 2, 3]) >>> Permutation([1, 0, 2, 3], size=20) Permutation([1, 0], size=20) >>> Permutation([1, 0, 2, 4, 3, 5, 6], size=20) Permutation([1, 0, 2, 4, 3], size=20)
>>> p = Permutation([1, 0, 2, 3]) >>> Permutation.print_cyclic = True >>> p (3)(0 1) >>> Permutation.print_cyclic = False
The 2 was not printed but it is still there as can be seen with the array_form and size methods:
>>> p.array_form [1, 0, 2, 3] >>> p.size 4
Short Introduction To Other Methods
The permutation can act as a bijective function, telling what element is located at a given position
>>> q = Permutation([5, 2, 3, 4, 1, 0]) >>> q.array_form[1] # the hard way 2 >>> q(1) # the easy way 2 >>> {i: q(i) for i in range(q.size)} # showing the bijection {0: 5, 1: 2, 2: 3, 3: 4, 4: 1, 5: 0}
The full cyclic form (including singletons) can be obtained:
>>> p.full_cyclic_form [[0, 1], [2], [3]]
Any permutation can be factored into transpositions of pairs of elements:
>>> Permutation([[1, 2], [3, 4, 5]]).transpositions() [(1, 2), (3, 5), (3, 4)] >>> Permutation.rmul(*[Permutation([ti], size=6) for ti in _]).cyclic_form [[1, 2], [3, 4, 5]]
The number of permutations on a set of n elements is given by n! and is called the cardinality.
>>> p.size 4 >>> p.cardinality 24
A given permutation has a rank among all the possible permutations of the same elements, but what that rank is depends on how the permutations are enumerated. (There are a number of different methods of doing so.) The lexicographic rank is given by the rank method and this rank is used to increment a permutation with addition/subtraction:
>>> p.rank() 6 >>> p + 1 Permutation([1, 0, 3, 2]) >>> p.next_lex() Permutation([1, 0, 3, 2]) >>> _.rank() 7 >>> p.unrank_lex(p.size, rank=7) Permutation([1, 0, 3, 2])
The product of two permutations p and q is defined as their composition as functions, (p*q)(i) = q(p(i)) [R34].
>>> p = Permutation([1, 0, 2, 3]) >>> q = Permutation([2, 3, 1, 0]) >>> list(q*p) [2, 3, 0, 1] >>> list(p*q) [3, 2, 1, 0] >>> [q(p(i)) for i in range(p.size)] [3, 2, 1, 0]
The permutation can be ‘applied’ to any listlike object, not only Permutations:
>>> p(['zero', 'one', 'four', 'two']) ['one', 'zero', 'four', 'two'] >>> p('zo42') ['o', 'z', '4', '2']
If you have a list of arbitrary elements, the corresponding permutation can be found with the from_sequence method:
>>> Permutation.from_sequence('SymPy') Permutation([1, 3, 2, 0, 4])

array_form
¶ Return a copy of the attribute _array_form Examples ========
>>> from sympy.combinatorics.permutations import Permutation >>> Permutation.print_cyclic = False >>> p = Permutation([[2, 0], [3, 1]]) >>> p.array_form [2, 3, 0, 1] >>> Permutation([[2, 0, 3, 1]]).array_form [3, 2, 0, 1] >>> Permutation([2, 0, 3, 1]).array_form [2, 0, 3, 1] >>> Permutation([[1, 2], [4, 5]]).array_form [0, 2, 1, 3, 5, 4]

ascents
()[source]¶ Returns the positions of ascents in a permutation, ie, the location where p[i] < p[i+1]
See also
descents
,inversions
,min
,max
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([4, 0, 1, 3, 2]) >>> p.ascents() [1, 2]

atoms
()[source]¶ Returns all the elements of a permutation
Examples
>>> from sympy.combinatorics import Permutation >>> Permutation([0, 1, 2, 3, 4, 5]).atoms() {0, 1, 2, 3, 4, 5} >>> Permutation([[0, 1], [2, 3], [4, 5]]).atoms() {0, 1, 2, 3, 4, 5}

cardinality
¶ Returns the number of all possible permutations.
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([0, 1, 2, 3]) >>> p.cardinality 24

commutator
(x)[source]¶ Return the commutator of self and x:
~x*~self*x*self
If f and g are part of a group, G, then the commutator of f and g is the group identity iff f and g commute, i.e. fg == gf.
References
http://en.wikipedia.org/wiki/Commutator
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> Permutation.print_cyclic = False >>> p = Permutation([0, 2, 3, 1]) >>> x = Permutation([2, 0, 3, 1]) >>> c = p.commutator(x); c Permutation([2, 1, 3, 0]) >>> c == ~x*~p*x*p True
>>> I = Permutation(3) >>> p = [I + i for i in range(6)] >>> for i in range(len(p)): ... for j in range(len(p)): ... c = p[i].commutator(p[j]) ... if p[i]*p[j] == p[j]*p[i]: ... assert c == I ... else: ... assert c != I ...

commutes_with
(other)[source]¶ Checks if the elements are commuting.
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> a = Permutation([1, 4, 3, 0, 2, 5]) >>> b = Permutation([0, 1, 2, 3, 4, 5]) >>> a.commutes_with(b) True >>> b = Permutation([2, 3, 5, 4, 1, 0]) >>> a.commutes_with(b) False

cycle_structure
¶ Return the cycle structure of the permutation as a dictionary indicating the multiplicity of each cycle length.
Examples
>>> from sympy.combinatorics import Permutation >>> Permutation.print_cyclic = True >>> Permutation(3).cycle_structure {1: 4} >>> Permutation(0, 4, 3)(1, 2)(5, 6).cycle_structure {2: 2, 3: 1}

cycles
¶ Returns the number of cycles contained in the permutation (including singletons).
Examples
>>> from sympy.combinatorics import Permutation >>> Permutation([0, 1, 2]).cycles 3 >>> Permutation([0, 1, 2]).full_cyclic_form [[0], [1], [2]] >>> Permutation(0, 1)(2, 3).cycles 2

cyclic_form
¶ This is used to convert to the cyclic notation from the canonical notation. Singletons are omitted.
See also
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> Permutation.print_cyclic = False >>> p = Permutation([0, 3, 1, 2]) >>> p.cyclic_form [[1, 3, 2]] >>> Permutation([1, 0, 2, 4, 3, 5]).cyclic_form [[0, 1], [3, 4]]

descents
()[source]¶ Returns the positions of descents in a permutation, ie, the location where p[i] > p[i+1]
See also
ascents
,inversions
,min
,max
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([4, 0, 1, 3, 2]) >>> p.descents() [0, 3]

classmethod
from_inversion_vector
(inversion)[source]¶ Calculates the permutation from the inversion vector.
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> Permutation.print_cyclic = False >>> Permutation.from_inversion_vector([3, 2, 1, 0, 0]) Permutation([3, 2, 1, 0, 4, 5])

classmethod
from_sequence
(i, key=None)[source]¶ Return the permutation needed to obtain
i
from the sorted elements ofi
. If custom sorting is desired, a key can be given.Examples
>>> from sympy.combinatorics import Permutation >>> Permutation.print_cyclic = True
>>> Permutation.from_sequence('SymPy') (4)(0 1 3) >>> _(sorted("SymPy")) ['S', 'y', 'm', 'P', 'y'] >>> Permutation.from_sequence('SymPy', key=lambda x: x.lower()) (4)(0 2)(1 3)

full_cyclic_form
¶ Return permutation in cyclic form including singletons.
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> Permutation([0, 2, 1]).full_cyclic_form [[0], [1, 2]]

get_adjacency_distance
(other)[source]¶ Computes the adjacency distance between two permutations.
This metric counts the number of times a pair i,j of jobs is adjacent in both p and p’. If n_adj is this quantity then the adjacency distance is n  n_adj  1 [1]
[1] Reeves, Colin R. Landscapes, Operators and Heuristic search, Annals of Operational Research, 86, pp 473490. (1999)
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([0, 3, 1, 2, 4]) >>> q = Permutation.josephus(4, 5, 2) >>> p.get_adjacency_distance(q) 3 >>> r = Permutation([0, 2, 1, 4, 3]) >>> p.get_adjacency_distance(r) 4

get_adjacency_matrix
()[source]¶ Computes the adjacency matrix of a permutation.
If job i is adjacent to job j in a permutation p then we set m[i, j] = 1 where m is the adjacency matrix of p.
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation.josephus(3, 6, 1) >>> p.get_adjacency_matrix() Matrix([ [0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 0, 1], [0, 1, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [0, 0, 0, 1, 0, 0]]) >>> q = Permutation([0, 1, 2, 3]) >>> q.get_adjacency_matrix() Matrix([ [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1], [0, 0, 0, 0]])

get_positional_distance
(other)[source]¶ Computes the positional distance between two permutations.
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([0, 3, 1, 2, 4]) >>> q = Permutation.josephus(4, 5, 2) >>> r = Permutation([3, 1, 4, 0, 2]) >>> p.get_positional_distance(q) 12 >>> p.get_positional_distance(r) 12

get_precedence_distance
(other)[source]¶ Computes the precedence distance between two permutations.
Suppose p and p’ represent n jobs. The precedence metric counts the number of times a job j is preceded by job i in both p and p’. This metric is commutative.
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([2, 0, 4, 3, 1]) >>> q = Permutation([3, 1, 2, 4, 0]) >>> p.get_precedence_distance(q) 7 >>> q.get_precedence_distance(p) 7

get_precedence_matrix
()[source]¶ Gets the precedence matrix. This is used for computing the distance between two permutations.
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation.josephus(3, 6, 1) >>> p Permutation([2, 5, 3, 1, 4, 0]) >>> p.get_precedence_matrix() Matrix([ [0, 0, 0, 0, 0, 0], [1, 0, 0, 0, 1, 0], [1, 1, 0, 1, 1, 1], [1, 1, 0, 0, 1, 0], [1, 0, 0, 0, 0, 0], [1, 1, 0, 1, 1, 0]])

index
()[source]¶ Returns the index of a permutation.
The index of a permutation is the sum of all subscripts j such that p[j] is greater than p[j+1].
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([3, 0, 2, 1, 4]) >>> p.index() 2

inversion_vector
()[source]¶ Return the inversion vector of the permutation.
The inversion vector consists of elements whose value indicates the number of elements in the permutation that are lesser than it and lie on its right hand side.
The inversion vector is the same as the Lehmer encoding of a permutation.
See also
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([4, 8, 0, 7, 1, 5, 3, 6, 2]) >>> p.inversion_vector() [4, 7, 0, 5, 0, 2, 1, 1] >>> p = Permutation([3, 2, 1, 0]) >>> p.inversion_vector() [3, 2, 1]
The inversion vector increases lexicographically with the rank of the permutation, the ith element cycling through 0..i.
>>> p = Permutation(2) >>> while p: ... print('%s %s %s' % (p, p.inversion_vector(), p.rank())) ... p = p.next_lex() ... Permutation([0, 1, 2]) [0, 0] 0 Permutation([0, 2, 1]) [0, 1] 1 Permutation([1, 0, 2]) [1, 0] 2 Permutation([1, 2, 0]) [1, 1] 3 Permutation([2, 0, 1]) [2, 0] 4 Permutation([2, 1, 0]) [2, 1] 5

inversions
()[source]¶ Computes the number of inversions of a permutation.
An inversion is where i > j but p[i] < p[j].
For small length of p, it iterates over all i and j values and calculates the number of inversions. For large length of p, it uses a variation of merge sort to calculate the number of inversions.
References
[1] http://www.cp.eng.chula.ac.th/~piak/teaching/algo/algo2008/countinv.htm
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([0, 1, 2, 3, 4, 5]) >>> p.inversions() 0 >>> Permutation([3, 2, 1, 0]).inversions() 6

is_Empty
¶ Checks to see if the permutation is a set with zero elements
See also
Examples
>>> from sympy.combinatorics import Permutation >>> Permutation([]).is_Empty True >>> Permutation([0]).is_Empty False

is_Identity
¶ Returns True if the Permutation is an identity permutation.
See also
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([]) >>> p.is_Identity True >>> p = Permutation([[0], [1], [2]]) >>> p.is_Identity True >>> p = Permutation([0, 1, 2]) >>> p.is_Identity True >>> p = Permutation([0, 2, 1]) >>> p.is_Identity False

is_Singleton
¶ Checks to see if the permutation contains only one number and is thus the only possible permutation of this set of numbers
See also
Examples
>>> from sympy.combinatorics import Permutation >>> Permutation([0]).is_Singleton True >>> Permutation([0, 1]).is_Singleton False

is_even
¶ Checks if a permutation is even.
See also
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([0, 1, 2, 3]) >>> p.is_even True >>> p = Permutation([3, 2, 1, 0]) >>> p.is_even True

is_odd
¶ Checks if a permutation is odd.
See also
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([0, 1, 2, 3]) >>> p.is_odd False >>> p = Permutation([3, 2, 0, 1]) >>> p.is_odd True

classmethod
josephus
(m, n, s=1)[source]¶ Return as a permutation the shuffling of range(n) using the Josephus scheme in which every mth item is selected until all have been chosen. The returned permutation has elements listed by the order in which they were selected.
The parameter
s
stops the selection process when there ares
items remaining and these are selected by continuing the selection, counting by 1 rather than bym
.Consider selecting every 3rd item from 6 until only 2 remain:
choices chosen ======== ====== 012345 01 345 2 01 34 25 01 4 253 0 4 2531 0 25314 253140
References
 http://en.wikipedia.org/wiki/Flavius_Josephus
 http://en.wikipedia.org/wiki/Josephus_problem
 http://www.wou.edu/~burtonl/josephus.html
Examples
>>> from sympy.combinatorics import Permutation >>> Permutation.josephus(3, 6, 2).array_form [2, 5, 3, 1, 4, 0]

length
()[source]¶ Returns the number of integers moved by a permutation.
Examples
>>> from sympy.combinatorics import Permutation >>> Permutation([0, 3, 2, 1]).length() 2 >>> Permutation([[0, 1], [2, 3]]).length() 4

list
(size=None)[source]¶ Return the permutation as an explicit list, possibly trimming unmoved elements if size is less than the maximum element in the permutation; if this is desired, setting
size=1
will guarantee such trimming.Examples
>>> from sympy.combinatorics.permutations import Permutation >>> Permutation.print_cyclic = False >>> p = Permutation(2, 3)(4, 5) >>> p.list() [0, 1, 3, 2, 5, 4] >>> p.list(10) [0, 1, 3, 2, 5, 4, 6, 7, 8, 9]
Passing a length too small will trim trailing, unchanged elements in the permutation:
>>> Permutation(2, 4)(1, 2, 4).list(1) [0, 2, 1] >>> Permutation(3).list(1) []

max
()[source]¶ The maximum element moved by the permutation.
See also
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([1, 0, 2, 3, 4]) >>> p.max() 1

min
()[source]¶ The minimum element moved by the permutation.
See also
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([0, 1, 4, 3, 2]) >>> p.min() 2

next_lex
()[source]¶ Returns the next permutation in lexicographical order. If self is the last permutation in lexicographical order it returns None. See [4] section 2.4.
See also
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([2, 3, 1, 0]) >>> p = Permutation([2, 3, 1, 0]); p.rank() 17 >>> p = p.next_lex(); p.rank() 18

next_nonlex
()[source]¶ Returns the next permutation in nonlex order [3]. If self is the last permutation in this order it returns None.
See also
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> Permutation.print_cyclic = False >>> p = Permutation([2, 0, 3, 1]); p.rank_nonlex() 5 >>> p = p.next_nonlex(); p Permutation([3, 0, 1, 2]) >>> p.rank_nonlex() 6

next_trotterjohnson
()[source]¶ Returns the next permutation in TrotterJohnson order. If self is the last permutation it returns None. See [4] section 2.4. If it is desired to generate all such permutations, they can be generated in order more quickly with the
generate_bell
function.Examples
>>> from sympy.combinatorics.permutations import Permutation >>> Permutation.print_cyclic = False >>> p = Permutation([3, 0, 2, 1]) >>> p.rank_trotterjohnson() 4 >>> p = p.next_trotterjohnson(); p Permutation([0, 3, 2, 1]) >>> p.rank_trotterjohnson() 5

order
()[source]¶ Computes the order of a permutation.
When the permutation is raised to the power of its order it equals the identity permutation.
See also
identity
,cardinality
,length
,rank
,size
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> Permutation.print_cyclic = False >>> p = Permutation([3, 1, 5, 2, 4, 0]) >>> p.order() 4 >>> (p**(p.order())) Permutation([], size=6)

parity
()[source]¶ Computes the parity of a permutation.
The parity of a permutation reflects the parity of the number of inversions in the permutation, i.e., the number of pairs of x and y such that
x > y
butp[x] < p[y]
.See also
_af_parity
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([0, 1, 2, 3]) >>> p.parity() 0 >>> p = Permutation([3, 2, 0, 1]) >>> p.parity() 1

classmethod
random
(n)[source]¶ Generates a random permutation of length
n
.Uses the underlying Python pseudorandom number generator.
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> Permutation.random(2) in (Permutation([1, 0]), Permutation([0, 1])) True

rank
()[source]¶ Returns the lexicographic rank of the permutation.
See also
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([0, 1, 2, 3]) >>> p.rank() 0 >>> p = Permutation([3, 2, 1, 0]) >>> p.rank() 23

rank_nonlex
(inv_perm=None)[source]¶ This is a linear time ranking algorithm that does not enforce lexicographic order [3].
See also
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([0, 1, 2, 3]) >>> p.rank_nonlex() 23

rank_trotterjohnson
()[source]¶ Returns the Trotter Johnson rank, which we get from the minimal change algorithm. See [4] section 2.4.
See also
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([0, 1, 2, 3]) >>> p.rank_trotterjohnson() 0 >>> p = Permutation([0, 2, 1, 3]) >>> p.rank_trotterjohnson() 7

static
rmul
(*args)[source]¶ Return product of Permutations [a, b, c, …] as the Permutation whose ith value is a(b(c(i))).
a, b, c, … can be Permutation objects or tuples.
Notes
All items in the sequence will be parsed by Permutation as necessary as long as the first item is a Permutation:
>>> Permutation.rmul(a, [0, 2, 1]) == Permutation.rmul(a, b) True
The reverse order of arguments will raise a TypeError.
Examples
>>> from sympy.combinatorics.permutations import _af_rmul, Permutation >>> Permutation.print_cyclic = False
>>> a, b = [1, 0, 2], [0, 2, 1] >>> a = Permutation(a); b = Permutation(b) >>> list(Permutation.rmul(a, b)) [1, 2, 0] >>> [a(b(i)) for i in range(3)] [1, 2, 0]
This handles the operands in reverse order compared to the
*
operator:>>> a = Permutation(a); b = Permutation(b) >>> list(a*b) [2, 0, 1] >>> [b(a(i)) for i in range(3)] [2, 0, 1]

classmethod
rmul_with_af
(*args)[source]¶ same as rmul, but the elements of args are Permutation objects which have _array_form

runs
()[source]¶ Returns the runs of a permutation.
An ascending sequence in a permutation is called a run [5].
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([2, 5, 7, 3, 6, 0, 1, 4, 8]) >>> p.runs() [[2, 5, 7], [3, 6], [0, 1, 4, 8]] >>> q = Permutation([1,3,2,0]) >>> q.runs() [[1, 3], [2], [0]]

signature
()[source]¶ Gives the signature of the permutation needed to place the elements of the permutation in canonical order.
The signature is calculated as (1)^<number of inversions>
See also
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([0, 1, 2]) >>> p.inversions() 0 >>> p.signature() 1 >>> q = Permutation([0,2,1]) >>> q.inversions() 1 >>> q.signature() 1

size
¶ Returns the number of elements in the permutation.
See also
Examples
>>> from sympy.combinatorics import Permutation >>> Permutation([[3, 2], [0, 1]]).size 4

support
()[source]¶ Return the elements in permutation, P, for which P[i] != i.
Examples
>>> from sympy.combinatorics import Permutation >>> p = Permutation([[3, 2], [0, 1], [4]]) >>> p.array_form [1, 0, 3, 2, 4] >>> p.support() [0, 1, 2, 3]

transpositions
()[source]¶ Return the permutation decomposed into a list of transpositions.
It is always possible to express a permutation as the product of transpositions, see [1]
References
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> p = Permutation([[1, 2, 3], [0, 4, 5, 6, 7]]) >>> t = p.transpositions() >>> t [(0, 7), (0, 6), (0, 5), (0, 4), (1, 3), (1, 2)] >>> print(''.join(str(c) for c in t)) (0, 7)(0, 6)(0, 5)(0, 4)(1, 3)(1, 2) >>> Permutation.rmul(*[Permutation([ti], size=p.size) for ti in t]) == p True

classmethod
unrank_lex
(size, rank)[source]¶ Lexicographic permutation unranking.
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> Permutation.print_cyclic = False >>> a = Permutation.unrank_lex(5, 10) >>> a.rank() 10 >>> a Permutation([0, 2, 4, 1, 3])

classmethod
unrank_nonlex
(n, r)[source]¶ This is a linear time unranking algorithm that does not respect lexicographic order [3].
See also
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> Permutation.print_cyclic = False >>> Permutation.unrank_nonlex(4, 5) Permutation([2, 0, 3, 1]) >>> Permutation.unrank_nonlex(4, 1) Permutation([0, 1, 2, 3])


class
sympy.combinatorics.permutations.
Cycle
(*args)[source]¶ Wrapper around dict which provides the functionality of a disjoint cycle.
A cycle shows the rule to use to move subsets of elements to obtain a permutation. The Cycle class is more flexible than Permutation in that 1) all elements need not be present in order to investigate how multiple cycles act in sequence and 2) it can contain singletons:
>>> from sympy.combinatorics.permutations import Perm, Cycle
A Cycle will automatically parse a cycle given as a tuple on the rhs:
>>> Cycle(1, 2)(2, 3) (1 3 2)
The identity cycle, Cycle(), can be used to start a product:
>>> Cycle()(1, 2)(2, 3) (1 3 2)
The array form of a Cycle can be obtained by calling the list method (or passing it to the list function) and all elements from 0 will be shown:
>>> a = Cycle(1, 2) >>> a.list() [0, 2, 1] >>> list(a) [0, 2, 1]
If a larger (or smaller) range is desired use the list method and provide the desired size – but the Cycle cannot be truncated to a size smaller than the largest element that is out of place:
>>> b = Cycle(2, 4)(1, 2)(3, 1, 4)(1, 3) >>> b.list() [0, 2, 1, 3, 4] >>> b.list(b.size + 1) [0, 2, 1, 3, 4, 5] >>> b.list(1) [0, 2, 1]
Singletons are not shown when printing with one exception: the largest element is always shown – as a singleton if necessary:
>>> Cycle(1, 4, 10)(4, 5) (1 5 4 10) >>> Cycle(1, 2)(4)(5)(10) (1 2)(10)
The array form can be used to instantiate a Permutation so other properties of the permutation can be investigated:
>>> Perm(Cycle(1, 2)(3, 4).list()).transpositions() [(1, 2), (3, 4)]
See also
Notes
The underlying structure of the Cycle is a dictionary and although the __iter__ method has been redefined to give the array form of the cycle, the underlying dictionary items are still available with the such methods as items():
>>> list(Cycle(1, 2).items()) [(1, 2), (2, 1)]

list
(size=None)[source]¶ Return the cycles as an explicit list starting from 0 up to the greater of the largest value in the cycles and size.
Truncation of trailing unmoved items will occur when size is less than the maximum element in the cycle; if this is desired, setting
size=1
will guarantee such trimming.Examples
>>> from sympy.combinatorics.permutations import Cycle >>> from sympy.combinatorics.permutations import Permutation >>> Permutation.print_cyclic = False >>> p = Cycle(2, 3)(4, 5) >>> p.list() [0, 1, 3, 2, 5, 4] >>> p.list(10) [0, 1, 3, 2, 5, 4, 6, 7, 8, 9]
Passing a length too small will trim trailing, unchanged elements in the permutation:
>>> Cycle(2, 4)(1, 2, 4).list(1) [0, 2, 1]

Generators¶

generators.
symmetric
()¶ Generates the symmetric group of order n, Sn.
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> Permutation.print_cyclic = True >>> from sympy.combinatorics.generators import symmetric >>> list(symmetric(3)) [(2), (1 2), (2)(0 1), (0 1 2), (0 2 1), (0 2)]

generators.
cyclic
()¶ Generates the cyclic group of order n, Cn.
See also
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> Permutation.print_cyclic = True >>> from sympy.combinatorics.generators import cyclic >>> list(cyclic(5)) [(4), (0 1 2 3 4), (0 2 4 1 3), (0 3 1 4 2), (0 4 3 2 1)]

generators.
alternating
()¶ Generates the alternating group of order n, An.
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> Permutation.print_cyclic = True >>> from sympy.combinatorics.generators import alternating >>> list(alternating(3)) [(2), (0 1 2), (0 2 1)]

generators.
dihedral
()¶ Generates the dihedral group of order 2n, Dn.
The result is given as a subgroup of Sn, except for the special cases n=1 (the group S2) and n=2 (the Klein 4group) where that’s not possible and embeddings in S2 and S4 respectively are given.
See also
Examples
>>> from sympy.combinatorics.permutations import Permutation >>> Permutation.print_cyclic = True >>> from sympy.combinatorics.generators import dihedral >>> list(dihedral(3)) [(2), (0 2), (0 1 2), (1 2), (0 2 1), (2)(0 1)]